Problem: Simplify the following expression and state the condition under which the simplification is valid. $a = \dfrac{q^2 - 9}{q + 3}$
Explanation: First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = q$ $ b = \sqrt{9} = 3$ So we can rewrite the expression as: $a = \dfrac{({q} + {3})({q} {-3})} {q + 3} $ We can divide the numerator and denominator by $(q + 3)$ on condition that $q \neq -3$ Therefore $a = q - 3; q \neq -3$